Topological Closure of Subset is Subset of Topological Closure/Proof 1

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq K$ and $K \subseteq S$.

Then:

$\map \cl H \subseteq \map \cl K$

where $\map \cl H$ denotes the closure of $H$.

Proof

From Topological Closure is Closed, $\map \cl K$ is closed.

From Set is Subset of its Topological Closure:

$K \subseteq \map \cl K$

By Subset Relation is Transitive, it follows that:

$H \subseteq \map \cl K$

Hence, by definition of closure as the smallest closed set that contains $H$:

$\map \cl H \subseteq \map \cl K$

$\blacksquare$